置箱於木板上逐漸抬高木板的一端,當傾斜角37°時,此箱即開始下滑,並測得其下滑的加速度為g/5 (a)則箱與木板間的靜摩擦係數μs與動摩擦係數μk之比值μs/μk為何? (b)若繼續將木板一端抬高到傾斜角為53°,則箱下滑的加速度應為何?
竭誠歡迎您光臨本物理教學網站! 版主 張慶堂
置箱於木板上逐漸抬高木板的一端,當傾斜角37°時,此箱即開始下滑,並測得其下滑的加速度為g/5 (a)則箱與木板間的靜摩擦係數μs與動摩擦係數μk之比值μs/μk為何? (b)若繼續將木板一端抬高到傾斜角為53°,則箱下滑的加速度應為何?
Posted in 牛頓第二運動定律.
rev="post-28" No comments
– 2008 年 12 月 05 日
0 Responses
Stay in touch with the conversation, subscribe to the RSS feed for comments on this post.
You must be logged in to post a comment.